[Haskell-cafe] Test on identity?

[Dušan Kolář]

Well, it makes a difference for me if I have twice the same subtree or  sharing one subtree from several places.  Later, I add some markers, thus, I know it is already processed or not.

Adding unique numbers and counting them makes sense if I know the result. If not then I don't know how to exploit it. But I may be tired too much already. :-(

Anyway, probably more imperative style would be a better option.

Thanks all,

Dušan


8. července 2020 18:13:34 SELČ, Viktor Dukhovni <ietf-dane at dukhovni.org> napsal:
>On Wed, Jul 08, 2020 at 08:17:41AM +0200, Dušan Kolář wrote:
>
>> Nevertheless, I do even some transformations. Thus, I would like to
>know it is still a 
>> DAG, not adding, accidentally, a node.
>> 
>> Is there any way, if I have data like
>> 
>> data Ex
>>   = Val Int
>>   | Add Ex Ex
>>  
>> so that I can test that some value Val i === Val i ? I mean, the
>pointers go to the 
>> same data box? I could do that via some IORefs, AFAIK, but I don't
>think it is 
>> feasible. Maybe to tune the algorithm...
>
>If, for the same "n", two "distinct" leaf nodes "Val n" are possible,
>in
>what sense is what you have still a DAG?  If there's a difference
>between:
>
>        Add
>       /   \
>      /     \
>     /       \
>    v         v
>  Val 1     Val 1
>
>and:
>
>        Add
>       /   \
>      /     \
>      \    /
>       v  v
>       Val 1
>
>then perhaps the data model is flawed by failing to capture the
>distinguishing attributes of distinct leaf objects.  And of course
>you might also have:
>
>         Add
>        /   \
>       /     \
>       \    /
>        v  v
>        Add
>       /   \
>      /     \
>     /       \
>    v         v
>  Val 1     Val 2
>
>So the most portable approach would be to assign a unique serial number
>all the nodes, both "Add", and "Val", and check that there's only path
>from the root to each distinct node (by serial number).  Or,
>equivalently,
>a recursive enumeration of all the serial numbers contains no
>duplicates.
>
>-- 
>    Viktor.
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